A light bulb emits power of 800 , W. What is | vedclass

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(C) The intensity $I$ at a distance $r$ from a point source is given by $I = \frac{P}{4\pi r^2}$.Substituting the values: $I = \frac{800}{4 	imes 3.1

Vedclass · Accepted Answer

$2.09 	imes 10^{-7} \, T$

Vedclass · Answer

(C) The intensity $I$ at a distance $r$ from a point source is given by $I = \frac{P}{4\pi r^2}$.Substituting the values: $I = \frac{800}{4 	imes 3.14 	imes (3.5)^2} = \frac{800}{153.86} \approx 5.2 \, W/m^2$.The relation between intensity and electric field amplitude $E_m$ is $I = \frac{1}{2} \epsilon_0 c E_m^2$.$E_m = \sqrt{\frac{2I}{\epsilon_0 c}} = \sqrt{\frac{2 	imes 5.2}{8.854 	imes 10^{-12} 	imes 3 	imes 10^8}} \approx \sqrt{3912} \approx 62.55 \, V/m$.The maximum magnetic field $B_m$ is given by $B_m = \frac{E_m}{c}$.$B_m = \frac{62.55}{3 	imes 10^8} \approx 2.085 	imes 10^{-7} \, T \approx 2.09 	imes 10^{-7} \, T$.

$A$ light bulb emits power of $800 \, W$. What is the maximum value of the magnetic field at a distance of $3.5 \, m$ from the bulb?

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